$f(t) = -2t-7$ $g(n) = -4n^{3}-n^{2}+f(n)$ $h(t) = -5t^{3}-t^{2}-3t+6-2(f(t))$ $ h(f(-5)) = {?} $
First, let's solve for the value of the inner function, $f(-5)$ . Then we'll know what to plug into the outer function. $f(-5) = (-2)(-5)-7$ $f(-5) = 3$ Now we know that $f(-5) = 3$ . Let's solve for $h(f(-5))$ , which is $h(3)$ $h(3) = -5(3^{3})-3^{2}+(-3)(3)+6-2(f(3))$ To solve for the value of $h$ , we need to solve for the value of $f(3)$ $f(3) = (-2)(3)-7$ $f(3) = -13$ That means $h(3) = -5(3^{3})-3^{2}+(-3)(3)+6+(-2)(-13)$ $h(3) = -121$